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Solution.java
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Solution.java
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package _0015_3Sum;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
/**
* 使用查找表解决
* 时间复杂度:O(n^2)
* 空间复杂度:O(n)
*/
public class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
if (nums.length < 3) return res;
HashMap<Integer, Integer> counter = new HashMap<>(); // 统计各元素出现次数,查找表
for (int i = 0; i < nums.length; i++) {
if (!counter.containsKey(nums[i])) {
counter.put(nums[i], 1);
} else {
counter.put(nums[i], counter.get(nums[i]) + 1);
}
}
if (counter.containsKey(0) && counter.get(0) >= 3) {
res.add(generateList(0, 0, 0));
}
// 数组排序,默认升序排列
Arrays.sort(nums);
// 数组去重,返回去重后的新数组
nums = unique(nums);
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[i] * 2 + nums[j] == 0 && counter.get(nums[i]) >= 2)
res.add(generateList(nums[i], nums[i], nums[j]));
if (nums[i] + nums[j] * 2 == 0 && counter.get(nums[j]) >= 2)
res.add(generateList(nums[i], nums[j], nums[j]));
int c = -nums[i] - nums[j]; // 另一个需要寻找的元素
if (c > nums[j] && counter.containsKey(c))
res.add(generateList(nums[i], nums[j], c));
}
}
return res;
}
private List<Integer> generateList(int a, int b, int c) {
List<Integer> item = new ArrayList<>();
item.add(a);
item.add(b);
item.add(c);
return item;
}
// 对已经拍好序的数组去重,返回去重后的数组
// LeetCode 0026号问题
private int[] unique(int[] nums) {
if (nums.length == 0)
return nums;
int k = 1; // [0...k)中的元素均为去重后的元素
int removeValue = nums[0];
for (int i = 1; i < nums.length; i++)
if (nums[i] != removeValue) {
nums[k++] = nums[i];
removeValue = nums[i];
}
int[] newNums = new int[k];
for (int i = 0; i < newNums.length; i++)
newNums[i] = nums[i];
return newNums;
}
public static void main(String[] args) {
int[] nums = {-1, 0, 1, 2, -1, -4};
System.out.println((new Solution()).threeSum(nums)); // [[-1, 0, 1], [-1, -1, 2]]
}
}